CaS exists in a cubic close packed arrangement of `S^(2-)` ions in which `Ca^(2+)` ions occupy 1/2 of the available tetrahedral holes. How many `Ca^(2+)and S^(2-)` ions are contained in the unit cell?

Options

A. 1,1
B. 2,4
C. 4,4
D. 4,2

Answer

To determine the number of \( \text{Ca}^{2+} \) and \( \text{S}^{2-} \) ions in the unit cell of calcium sulfide (\( \text{CaS} \)), we follow these steps:

### Step 1: Understand the arrangement of ions
Calcium sulfide exists in a cubic close-packed (CCP) arrangement of \( \text{S}^{2-} \) ions. In a CCP structure, the ions are arranged in a face-centered cubic (FCC) lattice.

### Step 2: Calculate the number of \( \text{S}^{2-} \) ions in the unit cell
In a CCP arrangement, the total number of ions in the unit cell can be calculated as follows:
- There are 8 corner atoms, and each corner atom contributes \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, and each face atom contributes \( \frac{1}{2} \) to the unit cell.

Calculating the contributions:
- Contribution from corner atoms:
\[ 8 \text{ corners} \times \frac{1}{8} = 1 \text{ ion} \]
- Contribution from face-centered atoms:
\[ 6 \text{ faces} \times \frac{1}{2} = 3 \text{ ions} \]

Thus, the total number of \( \text{S}^{2-} \) ions in the unit cell is:
\[ 1 + 3 = 4 \text{ ions} \]

### Step 3: Calculate the number of tetrahedral holes
In a CCP structure, the number of tetrahedral holes is twice the number of atoms in the unit cell. Since we have 4 \( \text{S}^{2-} \) ions:
\[ \text{Total tetrahedral holes} = 2 \times 4 = 8 \text{ holes} \]

### Step 4: Determine the number of \( \text{Ca}^{2+} \) ions
It is given that \( \text{Ca}^{2+} \) ions occupy half of the available tetrahedral holes. Therefore, the number of \( \text{Ca}^{2+} \) ions is:
\[ \text{Number of } \text{Ca}^{2+} \text{ ions} = \frac{1}{2} \times 8 = 4 \text{ ions} \]

### Step 5: Final result
Thus, in one unit cell of \( \text{CaS} \):
- Number of \( \text{Ca}^{2+} \) ions = 4
- Number of \( \text{S}^{2-} \) ions = 4

### Conclusion
The final answer is:
\[ \text{Number of } \text{Ca}^{2+} \text{ ions} = 4, \quad \text{Number of } \text{S}^{2-} \text{ ions} = 4 \]

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